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Cauchy sequences

  1. Jun 16, 2005 #1
    hello all

    I found this rather interesting
    suppose that a sequence [tex]{x_{n}}[/tex] satisfies

    [tex] |x_{n+1}-x_{n}|<\frac{1}{n+1}[/tex] [tex] \forall n\epsilon N[/tex]

    how couldnt the sequence [tex]{x_{n}}[/tex] not be cauchy? I tried to think of some examples to disprove it but i didnt achieve anything doing that, please help

  2. jcsd
  3. Jun 16, 2005 #2


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    An obvious one: just add those differences. Let xn be [tex]\sum_{i=1}^n \frac{1}{n}[/tex]. That series does not converge and so the sequence of partial sums is not Cauchy.
  4. Jun 17, 2005 #3


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    I'm hesitant to challenge you Hall but feel compelled to do so at grave risk as Analysis is not my strong suite. Alright nothing is but I digress.

    It looks like a Cauchy sequence to me based on the definition:

    The sequence: [itex]\{x_n}\}[/itex] is a Cauchy sequence if given [itex]\epsilon[/itex] there exists N such that for all m,n[itex]\leq[/itex]N we have:


    For the sequence:

    [tex] |x_{n+1}-x_{n}|<\frac{1}{n+1}[/tex]

    I take [itex]\epsilon=\frac{1}{N+1}[/itex];

    Thus for all n>N:

  5. Jun 17, 2005 #4
    That's true; but look closely - is that really the same thing as

  6. Jun 17, 2005 #5


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    For a given &epsilon;, you then choose a large enough N. No fair choosing N first and then finding &epsilon; so that N doesn't work!
    You do remind me that just [tex]|x_{n+1}-x_n|< \epsilon[/tex] is not enough. You have to show that [tex]|x_n-x_m|< \epsilon[/tex] which is basically showing that the tail of the sum goes to 0.
  7. Jun 17, 2005 #6


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    I suppose not.


    does not imply


    Still a little unclear. I'll work on it. Thanks guys.

    Seriously, why do I even bother with differential equations anyway . . .
  8. Jun 17, 2005 #7
    Simply put: for given N, your

    [tex] | x_m - x_n | [/tex],

    [tex] \mbox{ as } m \rightarrow \infty [/tex],

    is not guaranteed to be any better than the (divergent) harmonic series:

    [tex] | x_m - x_n | < \sum_{k=n+1}^{\infty} \frac{1}{n} \rightarrow \infty [/tex]
  9. Jun 17, 2005 #8


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    Alright, I'm interested in getting my math right.
    Following Hall's example (a slight change):


    [tex]x_n=\sum_{i=1}^{n+1} \frac{1}{i}[/tex]


    [tex]|x_{n+1}-x_n|=|\sum_{n=1}^{n+2} \frac{1}{i} - \sum_{n=1}^{n+1} \frac{1}{i}|=\frac{1}{n+2}[/tex]



    However, the partial sums of this harmonic series [itex] \{x_n\}[/itex] do not converge and thus cannot be a Cauchy sequence.

    Now, I got a pot of spagetti to make . . .
  10. Jun 17, 2005 #9
    hello all

    nice stuff saltydog, i had a similar result, i think your fine with analysis, and hallsofivy your right it is an obvious one but the funny thing is i cant think of any other examples, do you know of any others?

  11. Jul 27, 2011 #10
    hi guys i need bounded but not cauchy function can you help me asap?
  12. Jul 28, 2011 #11
    Kevin, consider any bounded periodic function that is not constant (like, for instance, a sinusoid). It will never become and remain arbitrarily close to anything, but it does remain within fixed bounds.
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